Problem 40 - Chemistry: The Central Science

Question

Consider a transition of the electron in the hydrogen atom from \(n=4\) to \(n=9\). (a) Is \(\Delta E\) for this process positive or negative? (b) Determine the wavelength of light that is associated with this transition. Will the light be absorbed or emitted? (c) In which portion of the electromagnetic spectrum is the light in part (b)?

Step-by-Step Solution

Answer

(a) The energy change, \(\Delta E\), for the electron transition in the hydrogen atom is positive (0.68 eV). (b) The wavelength of light associated with this transition is \(182 nm\), and the light is absorbed during the transition. (c) The light belongs to the ultraviolet (UV) portion of the electromagnetic spectrum.

Step 1: (1) Calculate the energy change in the transition

First, we need to calculate the energy change in the electron transition. The energy change formula for the hydrogen atom is: \(\Delta E = -13.6 \frac{Z^2}{n^2} eV\) where \(Z\) is the atomic number (for hydrogen, \(Z=1\)), and \(n\) is the principal quantum number. The energy change for this transition is the difference between the energy levels at n=4 and n=9: \(\Delta E = E_{n=9} - E_{n=4}\) Now, we calculate the energy levels at n=4 and n=9: \(E_{n=4} = -13.6 \frac{1^2}{4^2} = -0.85 eV\) \(E_{n=9} = -13.6 \frac{1^2}{9^2} = -0.17 eV\) Now, we calculate the energy change: \(\Delta E = E_{n=9} - E_{n=4} = -0.17 - (-0.85) = 0.68 eV\)

Step 2: (2) Determine if the energy change is positive or negative

The energy change, \(\Delta E\), is positive (0.68 eV). Therefore, the answer to part (a) is that the energy change is positive.

Step 3: (3) Calculate the wavelength of light associated with this transition

To find the wavelength of light associated with this transition, we should first convert the energy change from electron volts (eV) to Joules (J). We know that 1 eV = \(1.6 \times 10{-19} J\), so: \(\Delta E = 0.68 eV \times 1.6 \times 10^{-19} \frac{J}{eV} = 1.09 \times 10^{-19} J\) Now, we will use the formula relating the energy of a photon to its wavelength: \(E = \frac{hc}{\lambda}\) where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.63 \times 10^{-34} J \cdot s\)), \(c\) is the speed of light (\(3.00 \times 10^8 m/s\)), and \(\lambda\) is the wavelength of light. We want to find the wavelength, so we rearrange the formula for \(\lambda\): \(\lambda = \frac{hc}{E}\) Now, we plug in the values and calculate the wavelength: \(\lambda = \frac{6.63 \times 10^{-34} Js \cdot 3.00 \times 10^8 m/s}{1.09 \times 10^{-19} J} = 1.82 \times 10^{-7} m\)

Step 4: (4) Determine if the light is absorbed or emitted

Since the energy change is positive (\(\Delta E > 0\)), the electron gains energy. Therefore, the light is absorbed during this transition.

Step 5: (5) Identify the part of the electromagnetic spectrum

The wavelength we calculated is \(1.82 \times 10^{-7} m = 182 nm\), which falls in the ultraviolet (UV) region of the electromagnetic spectrum. To summarize: (a) The energy change, \(\Delta E\), is positive. (b) The wavelength of the light is \(182 nm\), and the light is absorbed. (c) The light belongs to the ultraviolet (UV) portion of the electromagnetic spectrum.

Key Concepts

Energy Change in Electron Transitions

When we think about electron transitions in a hydrogen atom, we're discussing the movement of an electron between different orbits or energy levels around the nucleus. An easy way to visualize this is to imagine an electron jumping between steps of a ladder where each step is a particular energy level.

Each energy level in an atom is characterized by a specific amount of energy; hence, as electrons transition from one level to another, they either absorb or release energy, resulting in an energy change, denoted as \(\Delta E\). In our example, an electron transition from the 4th to the 9th energy level in a hydrogen atom will naturally involve an energy change.

Applying the formula \(\Delta E = -13.6 \frac{Z^2}{n^2} eV\), where \(Z\) is the atomic number of the hydrogen atom, we find that the energy change is positive, indicating that energy is absorbed to promote the electron to a higher energy level. This absorbed energy corresponds to the energy of the light that would induce this transition.

Wavelength of Light in Transitions

The wavelength of light is intimately connected to the energy transitions of electrons. According to the formula \(E = \frac{hc}{\lambda}\), where \(E\) stands for the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength, we can see that when we know the energy involved in an electron transition, we can determine the wavelength of light associated with that transition.

In our exercise, the positive energy change of 0.68 eV is the key to finding the wavelength. First, we convert electron volts to Joules (since physics often prefers SI units), and then we rearrange our equation to solve for \(\lambda\). We find that the wavelength for this kind of transition is \(1.82 \times 10^{-7} m\), which means the light that would be absorbed in this process has a wavelength within this measure. It's quite spectacular to realize that each electron jump corresponds to a very specific color or kind of light!

Electromagnetic Spectrum Regions

The electromagnetic spectrum is a vast arena where all possible wavelengths of electromagnetic radiation reside. It's a whole spectrum of possibilities – from gamma rays, which have the shortest wavelengths, to radio waves, which stretch out to the longest wavelengths.

For our scenario, the wavelength of \(182 nm\) rests comfortably in the ultraviolet (UV) region of this spectrum. The UV region lies between visible light, which our eyes can detect, and X-rays, which are used in medical imaging. Understanding the regions of the electromagnetic spectrum not only helps us to categorize the type of radiation associated with electron transitions but also to appreciate the diverse applications and phenomena that are related to different types of electromagnetic waves. For example, while UV rays can be used to sterilize equipment, infrared radiation is notable for its use in thermal imaging.